Given : Thickness of slabs s1, s2, s3, s4, h=160 mm, s6: h=210 mm and covering load g_{επ}=2.0 kN/m^{2}. Question : Perform static analysis to determine the distribution of slabs loads transferred onto beams.
Figure 4.9.71 Figure 4.9.71
g_{1,2,3,4,5}=0.16x1.0x25.0=4.0 kN/m^{2}, g_{6}=0.21x1.0x25.0=5.25 kN/m^{2}, g_{επ}=2.0 kN/m^{2} Analysis will be performed using the global load (due to the small value of the live load): p_{1,2,3,4,5}=6.0x1.35+2.0x1.5=11.1 kN/m^{2}, p_{6}=7.25x1.35+2.0x1.5=12.80 kN/m^{2} Thus, on 1.0 m wide strips, the respective loads are p_{1, 2,3,4,5}=11.1 kN/m and p_{6}=12.80 kN/m. Slab s1
Figure 4.9.72 Figure 4.9.72
The slab is fixed at one edge:
From table 502, by switching indices x, y, we get:
From table 49, by switching indices x, y, we get:
Figure 4.9.75 Figure 4.9.75
Figure 4.9.76 Figure 4.9.76
The slab is supported on three edges. From table 55, by switching indices x, y, we get:
Figure 4.9.77 Figure 4.9.77
The slab is supported on two adjacent edges. From table 59 we get:
Figure 4.9.711 Figure 4.9.711
· In general, the most unfavourable span moments, arise in oneway slabs (s1) or cantilever slabs (s4) at one point in the first direction and at one line in the other. In twoway slabs moments are developed at the same point in both directions. · The moment peak values decrease rapidly with the distance from the peak, while at the fixed supports of slabs supported on three or two adjacent edges the decrease is quite rapid. · The moment at the support s5s6 or at the beam support of s6, in the middle of its 1.00 m wide strip, where the design calculations will be performed, may be much lesser. To this end, the reinforcement calculated in these supports will decrease significantly with the distance from supports.
Figure 4.9.79 Figure 4.9.79
Figure 4.9.710 Figure 4.9.710
Regarding slab s1 and assuming load distribution at an angle of 45^{ο} or 60^{ο} we have:
For a bar fixed at one end the resulted quotient is:
which is almost the same with the quotient resulted from the assumption of angles. Slab loads are distributed onto beams according to their influence region. The loading (kN/m) at each point of the beam is equal to the load ( kN/m^{2}) multiplied by the respective abscissa (m). Example : Loads on beams b4, b5
Figure 4.9.711 Figure 4.9.711
The triangular influence region of slab s5 load has an area of 1.53x2.65/2=2.03 m ^{2 } resulting to load of 2.03m ^{2} x11.1kN/m ^{2} =22.53 kN and eventually to equivalent uniform load of 22.53kN/2.65m= 8.5 kN/m . From slab s1: due to the rectangular influence region, equivalent uniform load directly results to 2.22mx11.1kN/m^{2}=24.6 kN/m. From slab s2: trapezoidal influence region has an area of (6.20+2.49)x2.35/2=10.21 m^{2} which gives load of 10.21m^{2}x11.1kN/m ^{2}=113.33kN and finally equivalent uniform load of 113.33kN/6.20m=18.3 kN/m. Overall: equivalent uniform load of beam b5 = 24.6 + 18.3=42.9 kN/m Equivalent uniform loadings are obtained directly by tables. For example, from table 502 for the distribution of load transferred from slab s2 onto b5, it yields: p_{yerm}=υ_{x}_{erm} · p · l_{y} =0.35x11.1x4.70=18.3 kΝ/m, while from table 55 for the distribution of load transferred from slab s5 onto b4, it yields: p_{yr}=υ_{xr} · p x l_{·} =0.29x11.1x2.65=8.5 kΝ/m.
Figure 4.9.712: Bending moment distribution, produced by the related software Figure 4.9.712: Bending moment distribution, produced by the related software
